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I will first give a short explanation of what I am asking, and then give a more comprehensive context.

If we have a LTI dynamic system acted upon by inputs $y(t)$ and producing outputs $x(t)$,

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we know that we can calculate a response to any arbitrary input if we know the system's unit impulse response $w(t)$, i.e. its response to a unit impulse input, or Dirac function, $y(t)=\delta(t)$. But what if the system is such that its inputs are differentiated, so that the impulse we feed into it gets differentiated as well? Would the corresponding response still be considered impulse response and be applicable for obtaining solutions to arbitrary inputs?

To further explain let's look at an example. Say we model a mechanical system and obtain the following LTI ordinary differential equation:

$m\ddot{x} + b\dot{x} + cx = y$.

To get $x(t)=w(t)$, we would set $y(t) = \delta(t)$ and solve either directly and get $w(t)$ or use Laplace transform and get $W(s)$. We would then use $w(t)$ or $W(s)$ to calculate the output to arbitrary input $y(t)$, either directly using convolution or using mutiplication in s-domain (and then inverse Laplace transform of $X(s)$ if we really need $x(t)$, which we often don't):

$x(t) = w(t)*y(t)$

$X(s) = W(s)Y(s)$.

$W(s)$ is the transfer function of the system (not proving this here, just stating), so let's focus on the second option ($s$-domain) because of the advantages of Laplace transform approach, such as algebraic simplicity and gaining insight about system response from $W(s)$ alone.

Now, what if our system was just a little different, like below (this is not contrived, for instance it corresponds to a simple automobile suspension model)?

$m\ddot{x} + b\dot{x} + cx = b\dot{y}$

The input gets differentiated! If we now set input $y(t) = \delta(t)$, it will act on the system differentiated. So although unit impulse is the input to the system, when we compare what we get on right hand sides of this and the first ODE, it does not quite seem straigtforward to call the response impulse response. However, there is no formal reason not to either. Let's make this more general to illustrate.

Using polynomial differential operators $P(D)$ and $Q(D)$, a general form of a LTI ODE is:

$P(D)x = Q(D)y$

In both above examples, $P(D)=mD^2 + bD + c$, while $Q(D) = 1$ in the first example and $Q(D) = bD$ in the second example. You can see that $D$, $D^2$ and so on stand for differentiation - if we bear this in mind we can formally multiply $x$ and $y$ by polynomials of $D$, $P(D)$ and $Q(D)$, and use this simple and convenient notation to write any LTI ODE. So if we have a delta function as input in a general LTI ODE, we can write this as follows:

$P(D)x = Q(D)y, y = \delta(t)$

Usually when impulse response is considered, only the case $P(D)x=y$ is explicitly considered, i.e. it is assumed $Q(D)=1$ so we have strictly $\delta(t)$ on the right hand side. However I see no formal or practical reason why we wouldn't admit the general form of the ODE as above and still draw exactly the same conclusions about solving ODEs with arbitrary inputs $f(t)$ using the impulse response $w(t)$. Specifically, I think we could say that for

$P(D)x = Q(D)f(t), f(t) \text{arbitrary, rest I.C.}, $

the solution is given by a convolution

$x(t)=w(t)*f(t)$,

which can also be calculated as its Laplace transform

$X(s)=W(s)F(s), W(s)=\frac {Q(s)}{P(s)}$,

where $W(s)$ is the transfer function, i.e. the Laplace transform of the unit impulse response $w(t)$ of the given ODE (where $Q(D)$ actually contains some derivatives). Is this correct?

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  • $\begingroup$ the W is modified to include the differential. $\endgroup$ – Stanley Pawlukiewicz Sep 22 '18 at 18:38
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I think your question can be reduced to the simpler question:

Is a differentiator fully characterized by its transfer function?

which is equivalent to asking

Is a differentiator a linear time-invariant (LTI) system?

The answer to the latter question is easily shown to be in the affirmative.

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Having read your terse answers, I have derived a simple formal proof that the answer to my above question is yes.

Given a system represented by an LTI ODE with rest (zero) initial conditions, of the form:

$P(D)x=f(t), \text{rest I.C.,}$

it is a fact that we can calculate the response to arbitrary input $f(t)$ if we know the impulse response $w(t)$, i.e. the solution $x = w(t)$ to :

$P(D)x=\delta(t) \text{, rest I.C.,}$

as the following convolution or multiplication in the s-domain:

$x(t) = w(t)*f(t)$

$X(s) = W(s)*F(s)$.

The proof rests on the fact that the system is indeed linear, time invariant, in other words that a linear combination of several time-shifted inputs leads to the same linear combination of equally time-shifted outputs to those inputs when they act alone and without time shift. Therefore, to prove that this also holds for a general LTI ODE:

$P(D)x=Q(D)f(t), \text{rest I.C.,}$

one must only prove that this is also a LTI system. We know that the left hand side is LTI, so we only need to prove that the right hand side is as well, i.e. that $Q(D)$ acting on a linear combination of several time shifted functions gives the same linear combination of equally time-shifted functions that are each individually acted on by $Q(D)$. Since the right hand side is of the exact same form as the left one, it obviously also is LTI. Therefore, the above methods for computing outputs to arbitrary inputs must also be valid. The output to impulse input $f(t) = \delta(t)$ can indeed be considered as the impuse response $w(t)$ and used in the convolution or transformed to obtain the transfer function $W(s)$.

This completes the proof, but I will add a short explicit demonstration that $Q(D)f(t)$ is linear, time invariant.

Proof of linearity of $Q(D)f(t)$

Linearity means that if the input is a linear combination of two inputs $y_1$ and $y_2$, i.e. $f(t) = ay_1+by_2$, after it gets operated on (differentiated) by $Q(D)$ it must become the same combination of individually differentiated inputs, i.e. $aQ(D)y_1+bQ(D)y_2$.

Since $Q(D)$ consists only of a lot of derivatives, a linear combination of these inputs will indeed give:

$Q(D)(ay_1+by_2)=aQ(D)y_1+bQ(D)y_2$

So indeed $Q(D)f(t)$ is linear.

Proof of time invariance of $Q(D)f(t)$

Time invariance means that if the input is first time shifted by $\tau$ to be $f(t-\tau)$ and only then gets operated on (differentiated) by $Q(D)$, it must give the same solution as when the non-shifted input $f(t)$ is first differentiated, and only then the result time shifted by $\tau$.

This is impractical to prove using $Q(D)$ notation, so lets show it for individual terms $a_kD^k$ of $Q(D) = a_nD^n + a_{n-1}D^{n-1} + ... + a_1D + a_0$, where the $a_k$s are constants and $D^k$ is $\frac{d^k}{dt^k}$.

Use substitution:

$f(t-\tau) = f(t^*), t^* = t-\tau$

By chain rule:

$$\frac{d}{dt}f(t^*) = \frac{df(t^*)}{dt^*}\frac{dt^*}{dt} = \frac{df(t^*)}{dt^*}\frac{d}{dt}(t-\tau) = \frac{d}{dt^*}f(t^*)$$

Using alternative notation for leftmost and rightmost expressions above makes the order of operations clearer:

$[f(t^*)]' = f'(t^*)$

On the left, we first shift then differentiate, while on the right, we first differentiate then shift. The results are equal, so first order differentiation is time invariant. This easily propagates to higher derivatives, i.e. higher terms $a_kD^k$ of $Q(D)$, thus $Q(D)f(t)$ is indeed time invariant.

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